Examine the potentiometer circuit arranged as shown in the figure....

The total resistance of the primary circuit is the sum of the internal resistance of the battery and the resistance of the potentiometer wire:

$R_{\text{eq}} = r + R = r + 15r = 16r$

The current flowing through the primary circuit is:

$I = \dfrac{\mathcal{E}}{R_{\text{eq}}} = \dfrac{\mathcal{E}}{16r}$

The potential difference across the entire potentiometer wire of length $L$ is:

$V = I \times R = \left(\dfrac{\mathcal{E}}{16r}\right) \times 15r = \dfrac{15\mathcal{E}}{16}$

The potential gradient (potential drop per unit length) along the potentiometer wire is:

$k = \dfrac{V}{L} = \dfrac{15\mathcal{E}}{16L}$

For zero deflection in the galvanometer, the potential difference across the balancing length $x$ must be equal to the emf of the cell in the secondary circuit. Since no current flows through the galvanometer, the internal resistance $r$ of the secondary cell does not contribute to any potential drop.

$k x = \dfrac{\mathcal{E}}{2}$

Substituting the value of $k$:

\left(\dfrac{15\mathcal{E}}{16L}\right) x = \dfrac{\mathcal{E}}{2}$<br><br>$x = \dfrac{16L}{2 \times 15} = \dfrac{8L}{15}$<br><br>Given the total length of the potentiometer wire is$L = 600 \text{ cm}$:<br><br>$x = \dfrac{8 \times 600}{15} = 8 \times 40 = 320 \text{ cm}